Ask question

Ask question

All categories

4 years ago

11

PLEASE HELP I'M DESPERATE <3

2 answers:

Iteru [2.4K]4 years ago

6 0

Start by simplifying all complex terms (anything term that can be broken down into an easier thing to work with). 4^2 is equal to 4x4 or 16, and the square root of 25 is 5. Now we have an expression that looks like this:

16-25x2

Using PEMDAS, we continue simplifying by multiplying 5 and two together, and then subtracting this from 16.

16-5x2
16-10
6

The value of this expression is 6.

Tomtit [17]4 years ago

5 0

4 squared = 16
Square root of 25 = 5
Equation = 16-5x2
Do 5x2 first because of order of operations
16-10 = 6

You might be interested in

(2a)(2b) please help

DENIUS [597]

The answer is 2^a+b lol

3 0

3 years ago

Forty-two less than a number z

Tasya [4]

42 less then a number z....

z - 42 <== ur expression

4 0

3 years ago

Read 2 more answers

Evaluate the expression

zheka24 [161]

Answer:

Step-by-step explanation:

x³ = 4³ = 4 * 4 * 4 = 64

4 0

3 years ago

Read 2 more answers

Suppose the total area is represented by the equation x^2+14x+24. If the area 1 is x^2 square units and area IV is 24 units, wha

Nutka1998 [239]

Answer:

See Explanation

Step-by-step explanation:

Given

$Total = x^2 + 14x + 24$

$Area\ I = x^2$

$Area\ IV = 24$

Required

True about area II and III

The question is incomplete; so, I will answer using general knowledge.

If the total area is:

$Total = x^2 + 14x + 24$

And there are 4 areas, then;

$Total = Area\ I + Area\ II + Area\ III + Area\ IV$

Rewrite as:

$Total = Area\ I + Area\ IV+ Area\ II + Area\ III$

This gives:

$x^2 + 14x + 24 = x^2 + 24+ Area\ II + Area\ III$

Collect like terms

$x^2 -x^2 + 14x + 24 -24= Area\ II + Area\ III$

$14x = Area\ II + Area\ III$

Rewrite as:

$Area\ II + Area\ III = 14x$

<em>So, the sum of areas III and IV must be 14x</em>

4 0

3 years ago

1 2 3 4 5 6 7 8 9 10

sammy [17]

Step 1 should have been 27(6+30).

3 0

3 years ago